Mixing
For polynomial $g(x)$ satisfying $(g(a))^2+(g'(a))^2=0$, evaluate $lim_{xto...
$begingroup$
If $g(x)$ is a polynomial function and
$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$
Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$
means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$
So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$
with Squeeze Theorem, the limit must be equal to $1$.
But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$
Could some help me to explain it? Thanks.
limits
edited Jan 21 at 14:16
Blue
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
$endgroup$
|
show 1 more comment
$begingroup$
If $g(x)$ is a polynomial function and
$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$
Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$
means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$
So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$
with Squeeze Theorem, the limit must be equal to $1$.
But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$
Could some help me to explain it? Thanks.
limits
edited Jan 21 at 14:16
Blue
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
$endgroup$
$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
1
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19
|
show 1 more comment
$begingroup$
If $g(x)$ is a polynomial function and
$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$
Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$
means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$
So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$
with Squeeze Theorem, the limit must be equal to $1$.
But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$
Could some help me to explain it? Thanks.
limits
edited Jan 21 at 14:16
Blue
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
$endgroup$
If $g(x)$ is a polynomial function and
$$(g(alpha))^2+(g'(alpha))^2=0$$ then evaluate
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)}bigglfloor frac{g'(x)}{g(x)}biggrfloor $$
Try: from
$$(g(alpha))^2+(g'(alpha))^2=0quadimpliesquad g(alpha)=g'(alpha) = 0 tag{1}$$
means polynomial $g(x)=0$ has a repeated root, $x=alpha$.
Using
$$frac{g'(x)}{g(x)}-1<bigglfloorfrac{g'(x)}{g(x)}biggrfloor leq frac{g'(x)}{g(x)} tag{2}$$
So
$$lim_{xrightarrow alpha}bigg(frac{g'(x)}{g(x)}-1bigg)frac{g(x)}{g'(x)}<lim_{xrightarrow alpha}bigglfloorfrac{g'(x)}{g(x)}biggrfloor frac{g(x)}{g'(x)}leq lim_{xrightarrow alpha}frac{g'(x)}{g(x)} frac{g(x)}{g'(x)} tag{3}$$
with Squeeze Theorem, the limit must be equal to $1$.
But I have a doubt for left side how can I prove
$$displaystyle lim_{xrightarrow alpha}frac{g(x)}{g'(x)} = 0 tag{4}$$
Could some help me to explain it? Thanks.
limits
limits
edited Jan 21 at 14:16
Blue
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
edited Jan 21 at 14:16
Blue
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
edited Jan 21 at 14:16
Blue
48.8k870156
edited Jan 21 at 14:16
Blue
48.8k870156
edited Jan 21 at 14:16
Blue
48.8k870156
48.8k870156
asked Jan 21 at 14:03
DXTDXT
5,9742732
asked Jan 21 at 14:03
DXTDXT
5,9742732
asked Jan 21 at 14:03
DXTDXT
5,9742732
5,9742732
$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
1
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19
|
show 1 more comment
$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
1
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19
$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
1
1
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$
You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$
if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.
All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.
But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
answered Jan 21 at 14:27
ShaqShaq
3049
$endgroup$
add a comment |
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1 Answer
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$begingroup$
$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$
You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$
if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.
All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.
But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
answered Jan 21 at 14:27
ShaqShaq
3049
$endgroup$
add a comment |
$begingroup$
$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$
You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$
if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.
All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.
But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
answered Jan 21 at 14:27
ShaqShaq
3049
$endgroup$
add a comment |
$begingroup$
$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$
You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$
if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.
All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.
But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
answered Jan 21 at 14:27
ShaqShaq
3049
$endgroup$
$lim_{xrightarrow a} frac{g(x)}{g'(x)}=0$
You can apply L'Hospital's Rule, and get: $ lim_{xrightarrow a} frac{g(x)}{g'(x)}=lim_{xrightarrow a} frac{g'(x)}{g''(x)}$
if $g''(a)neq 0$ you are done, otherwise you can keep going until you get $ lim_{xrightarrow a} frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{xrightarrow a} frac{g^{(n)}}{c}=frac{0}{c}$ when $c neq 0$.
All this is true if $g^{(k)}(x)neq 0$ for every $x neq a$ around $a$.
But let say for example $forall xin Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
answered Jan 21 at 14:27
ShaqShaq
3049
edited Jan 21 at 14:38
answered Jan 21 at 14:27
ShaqShaq
3049
answered Jan 21 at 14:27
ShaqShaq
3049
answered Jan 21 at 14:27
ShaqShaq
3049
3049
add a comment |
add a comment |
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$begingroup$
Do you work only with real numbers?
$endgroup$
– Mindlack
Jan 21 at 14:08
$begingroup$
yes mindlack...
$endgroup$
– DXT
Jan 21 at 14:11
$begingroup$
Recall that $g(x)=g(alpha)+(x-alpha)g'(x)$ by MVT. Can you take it from here?
$endgroup$
– learner
Jan 21 at 14:14
$begingroup$
Okay, what is an antiderivative $H$ for $G=frac{g’}{g}$? What kind of behavior has $H$ in a neighborhood of $alpha$? And what kind of behavior can $G$ have around $alpha$ that is compatible with the behavior of $H$? Alternatively, you can try and use Taylor expansion around $alpha$ to get equivalents for $g’(x)$ and $g(x)$.
$endgroup$
– Mindlack
Jan 21 at 14:16
1
$begingroup$
Learner: I am afraid your equation is wrong.
$endgroup$
– Mindlack
Jan 21 at 14:19